A nice approach to a famous sum.

[@BlackTigerClaws]

Hi, I don't know if there's a different definition of a "trapezoid" that I'm not aware of, but did you mean "tetrahedron"? Because as far as I'm aware, the volume formula you gave is exactly that for a tetrahedron.

[@divyanshaggarwal6243]

where i live, trapezoid generally refers to a trapezium. and yes he was referring to tetrahedrons

[@BoringExtrovert]

23:38 you had us in the first half not gonna lie

[@vaxjoaberg9452]

I've never heard of the word "trapezoid" meaning a 4-pointed, 3-dimensional object (which I've always understood to be a "tetrahedron"). I suppose this could be a case of linguistic or cultural confusion (eg, math v maths), but I think @michaelpenn is just committing a simple malapropism here. Whenever he says "trapezoid" just mentally replace it with "tetrahedron".

[@vaxjoaberg9452]

It's just a small part of the video so it's not a big deal.

[@marshallnoel2045]

Michael never ceases to amaze!!!

[@demenion3521]

just a note on the integral in the second tool: instead of integrating over x, y and z in this somewhat complicated way, we can also notice that the integral over any symmetric function of n variables in the region 0

[@dingo_dude]

this was very helpful review for a student in calculus 4 with a midterm this weekend. thanks for the great problem michael!

[@buxeessingh2571]

The transformation from the original tetrahedron to the reference one via T(u, v, w) is used in computer code of finite element computations.

[@jitzukinanaya4626]

The same substitution can as well be used to solve Basel problem by its 2-D case, and can be potentially generalized to n-D case to solve the even value of Riemann zeta function and odd value of Dirichlet beta function. At that time I learn about it, i think it's the most elegant thing in basic calculus. About the volume E', in u,v,w axes system, is actually two symmetric tetrahedrons joint with a common base of equilateral triangle, so it as six surface, we can call it bi-pyramid for it is symmetric against its common base. the point that 3 cutting surface joint, as the vertex of T2 in this video, and symmetric to the original point, is actually the center point of the basic cube of length π/2, this view may make more sense when calculating the volume of E'.

[@peterklenner2563]

3 blue 1 orange

[@electroskylightgaming4085]

Hol up

[@KaueMelo]

One of your best videos, for sure! :)

[@richardheiville937]

Euler in 18th century was already knowing the value of that sum. The change of variable used in the video was introduced by Beukers-Kolk-Calabi. A two dimensional version of this change of variable could be used to solve the Basel problem.

[@The1RandomFool]

This is definitely something I'd never conceive of. I used complex analysis to evaluate this series.

[@عمرانآلعمران-و7خ]

Hi Michael Your lectures are amazing! Could you please give problems involving hypergeometric series? Have a great day!

[@Reboxy1]

I actually solved this problem by turning it into an integral and using two results from your videos

[@goodplacetostop2973]

23:38 Good place to end this board 35:41 To all American reading this : while it’s important to stay involved in this election, remember to take care of your mental health during this period.

[@rupam6645]

Today your comment look odd.

[@wise_math]

Period meaning the pandemic, or the election?

[@megauser8512]

@@wise_math I think he meant election, since he didn't say pandemic, but he may have implied pandemic as well by saying health. However, we can only watch while they count the votes, but we can't vote anymore, since it is 3 days ****after**** election day, so it is out of our hands.

[@MTd2]

Give a like who wants the Rogers - Ramanujan identities series to continue!

[@marshallnoel2045]

What a beautiful mind blower!!!

[@BRUBRUETNONO]

Thanks for this nice journey

[@badremathsbadro7642]

Thank you

[@MrRyanroberson1]

Have you yet done many videos on gradients and/or contours? Scalar fields are an interesting topic to consider. Maybe there's a math problem about inverting one of these

[@MrRyanroberson1]

19:06 i would also notice the integral at that point is becoming 1/2 of (x-1)^2 by coincidence, so that becomes integral from -1 to 0 of x^2/2 dx -> 0-(-1)^3/6 = 1/6

[@daniellosh8341]

Just the corner points is not sufficient to determine the boundary of the solid. It remains to prove that the solid is bounded by flat planes under the curvilinear transformation.

[@juanixzx]

26:48 I think it's wrong in this statement 0

[@VaradMahashabde]

No they are separately true, the sum inequalities do not follow from the from variable-wise inequalities

[@juanixzx]

@@VaradMahashabde I suppose u, v, w is understood as "either u, v, w satisfy the inequality, and individually is true"

[@VaradMahashabde]

@@juanixzx I am not really sure what you mean precisely, so I'll simply right it in full. The following inequalities are individually true : 0 ≤ u ≤ π/2 0 ≤ v ≤ π/2 0 ≤ w ≤ π/2 0 ≤ u + v ≤ π/2 0 ≤ v + w ≤ π/2 0 ≤ w + u ≤ π/2

[@stewartcopeland4950]

@@VaradMahashabde u = 80, v = 9, w = 45 --> tan(u)*tan(v)*tan(w)= 0,89 < 1 but u+w = 125 > 90 !

[@tomasstride9590]

There is clearly something strange about the statement that u,v and w are all less than pi/2. Since tan of pi/2 is infinite the inequality is a bit strange. Your suggestion looks to me to be a better choice but I am unsure how to rigorously show it.

[@minwithoutintroduction]

مجهود كبير وعمل رائع . واصل

[@CM63_France]

Hi, For fun: 1 "so let's go ahead and write that down", 1 "so let's may be go ahead and write that down", 1 "so let's may be go ahead and do this", 1 "so may be let's go ahead and", 1 "may be we'll go ahead and write it like that", 1 "now let's go ahead and", 1 "let's go ahead and", 1 "so I'll go ahead and", 1 "so now we are going to go ahead and" 1 "I'll just like go ahead and", 1 "we can go ahead and" 1 "now we are going to go ahead and", 1 "great", 2 "ok, great", 1 "ok, sweet", 1 "now what I want to notice", 1 "so on and so forth".

[@michaelempeigne3519]

That is a tetrahedron, not a trapezoid @Michael_Penn

[@dudewaldo4]

Do you mean tetrahedron?

[@kqp1998gyy]

Awesome 💕

[@JalebJay]

When you were saying trapezoid, where can I find this definition? I only know it as the quadrilateral with a pair of parallel edges. Also, how do we find a deteriminate to a 3x1 matrix?

[@linggamusroji227]

we extend the matrix by (1,1,1) then calculate the determinant

[@BlackTigerClaws]

Since the 3 points he has listed in the matrix are in R3, they all have 3 coordinates, so the matrix is actually the matrix of those coordinates (e.g., A=(1,0,0), B=(1,1,0), C=(1,1,1), [[A],[B],[C]]=[[1,0,0],[1,1,0],[1,1,1]]), and then the determinant is calculated as usual.

[@riadsouissi]

notice that the integral substitution used here is known as Beukers-Kolk-Calabi.

[@lumpi806]

Nice ! Please, could you give me the name of the transformation whose the Jacobian is 1+x²y²z² ? Thanks !

[@alainrogez8485]

Guys, it is insane!

[@txikitofandango]

Trying to justify in some abstract n-dimensional sense how a trapezoid is equivalent to a tetrahedron haha. Anyway okay it was a good video, I really loved it!

[@VerSalieri]

Do you mean area of a trapezoid? Or volume of trapezoid by revolution? Can you please elaborate ? Trapezoids are planar objects, and thus do not posses a third dimension. Unless you are treating it as they would in certain aspects of physics where they mention linear volume (instead of length), 2 dimensional volumes and 3d volumes. Okay, nvm... I think what you call trapezoid we call a tetrahedron.

[@reamick]

He misspoke. He meant tetrahedron all along.

[@VerSalieri]

@@reamick He seplled it as trapezoiod and not trapezoid. This seems analogous to the naming of the paraboloid and ellipsoid.

[@tomatrix7525]

Epic

[@ccg8803]

Hi people, I really love this channel but actually I'm not able to understand all the staff it's explained in a proper way. Also I'm not an native english speaker so you'll find that to. The point is that, The jacobian transformations wants to be something like moving an 3D area from one place to another, No? And then the idea with this infinite sum is to transform it to a triple integral, which you've notice that, if you develop the correct transformation, you'll find a great cancellation in spide of deduce the sol'. Then, the second tool he use at the conclusion is like: Why he sum the area of the trapezoid below and then also the up-wards? As far as I'm concern it should be just the transformated Volum. But I guess I'm wrong. So, please, if someone somehow can help me it would be so apreciated.

[@shangaiguarisnaque9277]

I was thinking about translating Michael Penn's videos to Spanish (since that's my native language). Although I'm not sure if I can. I mean, I believe he has to allow translations in his channel

[@videolome]

Your transformation needs a +P1

[@holyshit922]

We can get this sum when we try to calculate integral Integral Int(ln^2(tan(x)),x=0..Pi/2) Substitution t=tan(x) , split the interval of integration [0..infinity] to [0..1] and [1..infinity] Substitute t = 1/u in integral on interval [1..infinity] You will get integral 2Int(ln^2(u)/(u^2+1),u=0..1) Expand 1/(u^2+1) to power seies Exchange order of summation and integration Integrate by parts twice to calculate integral Int(u^(2n)ln^2(u),u=0..1) Finally you should get 4sum((-1)^(n)/(2n+1)^3,n=0..infinity) and this sum is the sum from this video On the other hand Int(ln^2(tan(x)),x=0..Pi/2) can be calculated as follows Int(ln^2(tan(x)),x=0..Pi/2)=Int((ln(sin(x)/cos(x)))^2,x=0..Pi/2) =Int((ln(sin(x))-ln(cos(x)))^2,x=0..Pi/2) =Int(ln^2(sin(x)),x=0..Pi/2)-2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2)+Int(ln^2(cos(x)),x=0..Pi/2) =Int(ln^2(sin(Pi/2-t))*(-1),t=Pi..0)-2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2)+Int(ln^2(cos(x)),x=0..Pi/2) =Int(ln^2(cos(t)),t=0..Pi/2)-2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2)+Int(ln^2(cos(x)),x=0..Pi/2) =2(Int(ln^2(cos(x)),x=0..Pi/2) - Int(ln(sin(x))ln(cos(x)),x=0..Pi/2)) 4sum((-1)^(n)/(2n+1)^3,n=0..infinity) = 2(Int(ln^2(cos(x)),x=0..Pi/2) - Int(ln(sin(x))ln(cos(x)),x=0..Pi/2)) 2sum((-1)^(n)/(2n+1)^3,n=0..infinity) = Int(ln^2(cos(x)),x=0..Pi/2) - Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) And Michael calculated both of the integrals kzbin.info/www/bejne/n5zch3t7fdKahpo kzbin.info/www/bejne/oWHCdZqZl7CXo8U

[@lrosello]

I think that isn’t possible change the sum and the integration unless there is uniform convergence

[@fiartruck0125]

All y'all complaining that he called a tetrahedron a trapezoid, but I don't see any of you pedants complaining that he spelled it "trapezoiod"! :P

[@johnvandenberg8883]

Nice problem indeed Michael, but where did the absolute value in the Jacobian come from? It doesn't matter here, but in general it shouldn't be there.

[@Deegius]

tetrahedron !

[@GrmTrggr]

Wondering why he's not using the "Rule of Sarrus " to compute the determinant...

[@MA-bm9jz]

I personally never remember it,i usually try to make as many 0 s as i can before expanding

[@tonyhaddad1394]

You a proffesor ??? Man your level is my dream 💓💓

[@ericzeisel3522]

I don't understand all the complaints. Everyone knows t r a p e z o i o d is pronounced tet ruh hee drun, he just pronounced it wrong

[@frozenmoon998]

Please do the legend of Q6.. such a beautiful problem.

[@lost3834]

100th like!

[@blazedinfernape886]

Man i can't even understand what he is saying. Maybe in like 5years....

[@blazedinfernape886]

@Daniel Sam You took that very literally

[@wise_math]

Read many math books, start by elementary ones, until you understand what they say. Books usually give detail explanation. One reason why people don't understand math is bcos they are not used to it. Math is like a language.

[@makylemur7019]

As you provided no motivation or explanation for your approach to the problem what you did made absolutely no sense.

[@alexismiller2349]

Eh, it's just silly maths, sometimes it's fun to drop an out-of-the-blue solution. At least I think so